Question

Someone says that face-to-face classes always do better than online classes, you disagree and think they are the same. You gather data and below is some fictitious data.

Group 1	percentage	Group 2	percentage
Median	0.610	Median	0.74553571
Mean	0.646	Mean	0.68526786
St. Deviation	0.225	St. Deviation	0.18357571
n	16	n	18

Hint you'll need to convert the percentages to something else, and ask yourself are they proportion or mean...

Provide:

• Null and Alternative Hypothesis,
• Test statistic,
• p-value, (**alpha is 0.1),
• you need to determine whether we fail to reject or reject (aka test the claim),
• you'll also need to state it in non-technical terms.

Answer 1

Given that,
mean(x)=0.646
standard deviation , s.d1=0.225
number(n1)=16
y(mean)=0.685
standard deviation, s.d2 =0.183
number(n2)=18
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, α = 0.1
from standard normal table, two tailed t α/2 =1.753
since our test is two-tailed
reject Ho, if to < -1.753 OR if to > 1.753
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =0.646-0.685/sqrt((0.05063/16)+(0.03349/18))
to =-0.5502
| to | =0.5502
critical value
the value of |t α| with min (n1-1, n2-1) i.e 15 d.f is 1.753
we got |to| = 0.55019 & | t α | = 1.753
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -0.5502 ) = 0.59
hence value of p0.1 < 0.59,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: -0.5502
critical value: -1.753 , 1.753
decision: do not reject Ho
p-value: 0.59
we do not have enough evidence to support the claim that difference in means between two groups.